NCERT Solutions for Exercise 13.2 Class 9 Maths Chapter 13 - Surface Area and Volumes

Q1 The curved surface area of a right circular cylinder of height $14cm$ is $88c{m}^2$ . Find the diameter of the base of the cylinder.

Answer:

Given,

The curved surface area of the cylinder =

And, the height of the cylinder,

We know, Curved surface area of a right circular cylinder =

Therefore, the diameter of the cylinder =

Q2 It is required to make a closed cylindrical tank of height $1m$ and base diameter $140cm$ from a metal sheet. How many square metres of the sheet are required for the same?

Answer:

Given,

Height of the cylindrical tank = $h=1m$

Base diameter = $d=140cm=1.4m$

We know,

The total surface area of a cylindrical tank = $2\pi rh+2\pi r^2=2\pi r(r+h)$

$=2.\frac{22}{7}.\frac{1.4}{2}.(0.7+1)=2.\frac{22}{7}.(0.7).(1.7)$

$=44\times 0.17$

$=7.48m^2$

Therefore, square metres of the sheet is $7.48m^2$

Q3 (i) A metal pipe is $77cm$ long. The inner diameter of a cross section is $4cm$ , the outer diameter being $4.4cm$ (see Fig. $13.11$ ). Find its inner curved surface area,

Answer:

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder, $h=77cm$

Outer diameter = $r_1=4.4cm$

Inner diameter = $r_2=4cm$

Inner curved surface area = $2\pi r_{2}h$

$$\begin{gathered} =2\times \frac{22}{7}\times 2\times 77 \\ =968c{m}^2 \end{gathered}$$

Therefore, the inner curved surface area of the cylindrical pipe is $968c{m}^2$

Q3 (ii) A metal pipe is $77cm$ long. The inner diameter of a cross section is $4cm$ , the outerdiameter being $4.4cm$ (see Fig. $13.11$ ). Find its outer surface area.

Answer:

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder, $h=77cm$

Outer diameter = $r_1=4.4cm$

Inner diameter = $r_2=4cm$

Outer curved surface area = $2\pi r_{1}h$

$$\begin{gathered} =2\times \frac{22}{7}\times 2.2\times 77 \\ =1064.8c{m}^2 \end{gathered}$$

Therefore, the outer curved surface area of the cylindrical pipe is $1064.8c{m}^2$

Q3 (iii) A metal pipe is $77cm$ long. The inner diameter of a cross section is 4 cm, the outer diameter being $4.4cm$ (see Fig. $13.11$ ). Find its total surface area.

Answer:

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder, $h=77cm$

Outer diameter = $r_1=4.4cm$

Inner diameter = $r_2=4cm$

Outer curved surface area = $2\pi r_{1}h$

Inner curved surface area = $2\pi r_{2}h$

Area of the circular rings on top and bottom = $2\pi (r_2^2-r_1^2)$

$\therefore$ The total surface area of the pipe = $2\pi r_{1}h+2\pi r_{2}h+2\pi (r_2^2-r_1^2)$

$$\begin{gathered} =[968+1064.8+2\pi (2.2{)}^2-(2{)}^2] \\ =(2032.8+2\times \frac{22}{7}\times 0.84) \\ =(2032.8+5.28) \\ =2038.08c{m}^2 \end{gathered}$$

Therefore, the total surface area of the cylindrical pipe is $2038.08c{m}^2$

Q4 The diameter of a roller is $84cm$ and its length is $120cm$ . It takes $500$ complete revolutions to move once over to level a playground. Find the area of the playground in $m^2$ .

Answer:

Given,

The diameter of the cylindrical roller = $d=84cm$

Length of the cylindrical roller = $h=120cm$

The curved surface area of the roller = $2\pi rh=\pi dh$

$$\begin{gathered} =\frac{22}{7}\times 84\times 120 \\ =31680c{m}^2 \end{gathered}$$

$\therefore$ Area of the playground = $Areacoveredin1rotation\times 500$

$$\begin{gathered} =31680\times 500 \\ =15840000c{m}^2 \\ =1584m^2 \end{gathered}$$

Therefore, the required area of the playground = $1584m^2$

Q5 A cylindrical pillar is $50cm$ in diameter and $3.5m$ in height. Find the cost of painting the curved surface of the pillar at the rate of $Rs12.50$ per $m^2$ .

Answer:

Given,

Radius of the cylindrical pillar, r = $\frac{50}{2}cm=25cm=0.25m$

Height of the cylinder, h = $3.5m$

We know,

Curved surface area of a cylinder = $2\pi rh$

$\therefore$ Curved surface area of the pillar = $2\times \frac{22}{7}\times 0.25\times 3.5$

$=1\times 22\times 0.25m^2$

$=5.5m^2$

Now,

Cost of painting $1m^2$ of the pillar = $Rs12.50$

$\therefore$ Cost of painting the curved surface area of the pillar = $Rs.(12.50\times 5.5)$

$=Rs.68.75$

Therefore, the cost of painting curved surface area of the pillar is $Rs.68.75$

Q6 Curved surface area of a right circular cylinder is $4.4m^2$ . If the radius of the base of the cylinder is $0.7m$ , find its height.

Answer:

Given, a right circular cylinder

Curved surface area of the cylinder = $4.4m^2$

The radius of the base = $r=0.7m$

Let the height of the cylinder be $h$

We know,

Curved surface area of a cylinder of radius $r$ and height $h$ = $2\pi rh$

$\therefore$ $2\pi rh=4.4$

$$\begin{gathered} \Rightarrow 2\times \frac{22}{7}\times 0.7\times h=4.4 \\ \Rightarrow 44\times 0.1\times h=4.4 \\ \Rightarrow h=1m \end{gathered}$$

Therefore, the required height of the cylinder is $1m$

Q7 (i) The inner diameter of a circular well is $3.5m$ . It is $10m$ deep. Find its inner curved surface area.

Answer:

Given,

The inner diameter of the circular well = $d=3.5m$

Depth of the well = $h=10m$

We know,

The curved surface area of a cylinder = $2\pi rh$

$\therefore$ The curved surface area of the well = $2\times \frac{22}{7}\times \frac{3.5}{2}\times 10$

$$\begin{gathered} =44\times 0.25\times 10 \\ =110m^2 \end{gathered}$$

Therefore, the inner curved surface area of the circular well is $110m^2$

Q7 (ii) The inner diameter of a circular well is $3.5m$ . It is $10m$ deep. Find the cost of plastering this curved surface at the rate of Rs 40 per $m^2$ .

Answer:

Given,

The inner diameter of the circular well = $d=3.5m$

Depth of the well = $h=10m$

$\therefore$ The inner curved surface area of the circular well is $110m^2$

Now, the cost of plastering the curved surface per $m^2$ = Rs. 40

$\therefore$ Cost of plastering the curved surface of $110m^2$ = $Rs.(110\times 40)=Rs.4400$

Therefore, the cost of plastering the well is $Rs.4400$

Q8 In a hot water heating system, there is a cylindrical pipe of length $28m$ and diameter $5cm$ . Find the total radiating surface in the system.

Answer:

Given,

Length of the cylindrical pipe = $l=28m$

Diameter = $d=5cm=0.05m$

The total radiating surface will be the curved surface of this pipe.

We know,

The curved surface area of a cylindrical pipe of radius $r$ and length $l$ = $2\pi rl$

$\therefore$ CSA of this pipe =

$=2\times \frac{22}{7}\times \frac{0.05}{2}\times 28$

$$\begin{gathered} =22\times 0.05\times 4 \\ =4.4m^2 \end{gathered}$$

Therefore, the total radiating surface of the system is $4.4m^2$

Q9 (i) Find the lateral or curved surface area of a closed cylindrical petrol storage tank that is $4.2m$ in diameter and $4.5m$ high.

Answer:

Given, a closed cylindrical petrol tank.

The diameter of the tank = $d=4.2m$

Height of the tank = $h=4.5m$

We know,

The lateral surface area of a cylinder of radius $r$ and height $h$ = $2\pi rh$

$\therefore$ The lateral surface area of a cylindrical tank = $2\times \frac{22}{7}\times \frac{4.2}{2}\times 4.5$

$$\begin{gathered} =(44\times 0.3\times 4.5) \\ =59.4m^2 \end{gathered}$$

Therefore, the lateral or curved surface area of a closed cylindrical petrol storage tank is $59.4m^2$

Q9 (ii) Find: how much steel was actually used, if $\frac{1}{12}$ of the steel actually used was wasted in making the tank.

Answer:

Given, a closed cylindrical petrol tank.

The diameter of the tank = $d=4.2m$

Height of the tank = $h=4.5m$

Now, Total surface area of the tank = $2\pi r(r+h)$

$$\begin{gathered} =2\times \frac{22}{7}\times 2.1\times (2.1+4.5) \\ =(44\times 0.3\times 6.6) \\ =87.12m^2 \end{gathered}$$

Now, let $x{m}^2$ of steel sheet be actually used in making the tank

Since $\frac{1}{12}$ of steel was wasted, the left $\frac{11}{12}$ of the total steel sheet was used to made the tank.
$\therefore$ The total surface area of the tank = $\frac{11}{12}\times (Totalareaofsteelsheet)$
$$\begin{gathered} \Rightarrow 87.12=\frac{11}{12}x \\ \Rightarrow x=\frac{87.12\times 12}{11} \\ \Rightarrow x=95.04m^2 \end{gathered}$$

Therefore, $95.04m^2$ of steel was actually used in making the tank.

Q10 In Fig. $13.12$ , you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of $20cm$ and height of $30cm$ . A margin of $2.5cm$ is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Answer:

Given, a cylindrical lampshade

The diameter of the base = $d=20cm$

Height of the cylinder = $30cm$

The total height of lampshade= $(30+2.5+2.5)=35cm$

We know,

Curved surface area of a cylinder of radius $r$ and height $h$ = $2\pi rh$

Now, Cloth required for covering the lampshade = Curved surface area of the cylinder

$$\begin{gathered} =2\times \frac{22}{7}\times 10\times 35 \\ =22\times 10\times 10=2200c{m}^2 \end{gathered}$$

Therefore, $2200c{m}^2$ cloth will be required for covering the lampshade.

Q11 The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius $3cm$ and height $10.5cm$ . The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Answer:

Given, a cylinder with a base.

The radius of the cylinder = $r=3cm$

Height of the cylinder = $h=10.5cm$

We know,

The lateral surface area of a cylinder of radius $r$ and height $h$ = $2\pi rh$

$\therefore$ Area of the cylindrical penholder = Lateral areal + Base area

$=2\pi rh+\pi r^2=\pi r(2h+r)$

$=\frac{22}{7}\times 3\times [2(10.5)+3]$

$$\begin{gathered} =\frac{22}{7}\times 3\times [24] \\ =\frac{1584}{7}c{m}^2 \end{gathered}$$
Area of 35 penholders = $\frac{1584}{7}\times 35c{m}^2$

$=7920c{m}^2$

Therefore, the area of carboard required is $7920c{m}^2$